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21 Jun 2017, 01:07
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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64% (01:49) correct
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If -1＜x＜0, which of the following lists\(x^{-2}, x^{-1}, x, x^2, 1\) in an ascending order?
A. \(x^{-1}, x, x^2, 1, x^{-2}\)
B. \(x^{-2}, x, x^2, 1, x^{-1}\)
C. \(x^{-1}, x^2, x, 1, x^{-2}\)
D. \(x^{-1}, x, x^2, x^{-2}, 1\)
E. \(x, x^{-1}, x^2, 1, x^{-2}\)
A. \(x^{-1}, x, x^2, 1, x^{-2}\)
B. \(x^{-2}, x, x^2, 1, x^{-1}\)
C. \(x^{-1}, x^2, x, 1, x^{-2}\)
D. \(x^{-1}, x, x^2, x^{-2}, 1\)
E. \(x, x^{-1}, x^2, 1, x^{-2}\)
21 Jun 2017, 02:07
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Consider value of \(\frac{-1}{3}\) which falls in that range -1＜x＜0
Since we need the ascending order, we need to compute the values.
\(x^{-2} = (\frac{(-1)}{(3)})^{-2} = \frac{(-1)^{-2}}{(3)^{-2}} = 3^2 = 9\)
\(x^{-1} = (\frac{(-1)}{(3)})^{-1} = \frac{(-1)^{-1}}{(3)^{-1}} = -3\)
\(x^2 = (\frac{(-1)}{(3)})^{2} = \frac{(-1)^2}{(3)^2} =\frac{1}{9}\) = 0.1(approximately)
x = \(\frac{-1}{3}\) = -0.33(approximately)
Hence the ascending order of elements is \(x^{-1}, x, x^2, 1, x^{-2}\)(Option A)
Since we need the ascending order, we need to compute the values.
\(x^{-2} = (\frac{(-1)}{(3)})^{-2} = \frac{(-1)^{-2}}{(3)^{-2}} = 3^2 = 9\)
\(x^{-1} = (\frac{(-1)}{(3)})^{-1} = \frac{(-1)^{-1}}{(3)^{-1}} = -3\)
\(x^2 = (\frac{(-1)}{(3)})^{2} = \frac{(-1)^2}{(3)^2} =\frac{1}{9}\) = 0.1(approximately)
x = \(\frac{-1}{3}\) = -0.33(approximately)
Hence the ascending order of elements is \(x^{-1}, x, x^2, 1, x^{-2}\)(Option A)
23 Jun 2017, 01:08
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==> If you substitute x=-0.1, from \(x^{-1}=(-0.1)^{-1}=\frac{1}{(-0.1)}=-10, x=-0.1, x^2=(-0.1)^2=0.01, x^{-2}=(-0.1)^{-2}=\frac{1}{(-0.1)^2}=\frac{1}{0.01}=100\),
you get \(x^{-1}, x, x^2, 1, x^{-2}\).
The answer is A.
Answer: A
you get \(x^{-1}, x, x^2, 1, x^{-2}\).
The answer is A.
Answer: A
